Can we find the ratio of the areas of these two sectors?

Can we find the ratio of the areas of these two sectors?

Question

The figure shows two circles, centres $O$ and $C$, radii $r_1$ and $r_2$, which touch externally at $P$. Given that $A\hat{O}P=\dfrac{\pi}{4}$ radians, $B\hat{C}P = \dfrac{\pi}{3}$ radians and $AB$ is parallel to $OPC$, show that $\dfrac{r_1}{r_2} = \dfrac{\sqrt{6}}{2}$. Hence find the ratio of the areas of the sectors $OAP$, $CBP$.

Suggestion

Could we create some right-angled triangles within the figure, and use trigonometric functions here?

What’s the formula for the area of a sector?

Solution

Dropping perpendiculars from $A$ and $B$ gives us the above diagram. Thus we have

$$r_1 \sin \dfrac{\pi}{4} = AD= BE = r_2 \sin \frac{\pi}{3}.$$

And so we have $\dfrac{r_1}{r_2} = \dfrac{\sin \dfrac{\pi}{3}}{\sin \dfrac{\pi}{4}} = \dfrac{\left(\dfrac{\sqrt{3}}{2}\right)}{\left(\dfrac{1}{\sqrt{2}}\right)} = \dfrac{\sqrt{6}}{2}.$

The area of a sector containing an angle

$\theta$

and with radius

$r$

is

$\dfrac{1}{2}r^2 \theta$

. So

$$\begin{align*} A_1 &=\dfrac{1}{2}r_1^2 \times \dfrac{\pi}{4}=\dfrac{\pi r_1^2}{8},\\ A_2 &=\dfrac{1}{2}r_2^2 \times \dfrac{\pi}{3}=\dfrac{\pi r_2^2}{6}.\\ \end{align*}$$

Hence $\frac{A_1}{A_2} = \dfrac{6}{8} \left( \dfrac{r_1}{r_2} \right)^2 = \dfrac{9}{8}.$